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| author | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
| commit | 7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch) | |
| tree | 449d555969bfd7befe906877abab098c6e63a0e8 /3886/CH4/EX4.9/Ex4_9.sce | |
| parent | d1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff) | |
| download | Scilab-TBC-Uploads-master.tar.gz Scilab-TBC-Uploads-master.tar.bz2 Scilab-TBC-Uploads-master.zip | |
Diffstat (limited to '3886/CH4/EX4.9/Ex4_9.sce')
| -rw-r--r-- | 3886/CH4/EX4.9/Ex4_9.sce | 21 |
1 files changed, 21 insertions, 0 deletions
diff --git a/3886/CH4/EX4.9/Ex4_9.sce b/3886/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..59a686b2d --- /dev/null +++ b/3886/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,21 @@ +//Finding required forces
+//Refer fig. 4.16 (a)
+//Symmetry gives
+RE=(15+30+30+30+15)/2 //kN
+RA=RE //kN
+FAE=30 //kN
+//After construction as shown in ref. fig
+//Taking moment about C
+FAE=(60*5-15*5-30*2.5)/5 //kN (Tension)
+//assumptions are made as shown in fig. 4.16 (b)
+//Apply equilibrium conditions and solving equations
+FFC=15/0.366 //kN (Tension)
+FBC=(0.866*40.98+15)/0.707 //kN (Compression)
+//Lets analyse Joint B
+//Applying equilibrium conditions
+FBF=30*cosd(45) //kN (Compression)
+FAB=71.41+21.21 //kN (Compression)
+//Lets analyse Joint A
+//Applying equilibrium conditions
+FAF=(92.62*sind(45)-45)/sind(30) //kN (Tension)
+printf("\nThe required forces are:-\nFAB=%.2d kN (Compression)\nFBC=%.2d kN (Compression)\nFBF=%.2d kN (Compression)\nFAF=%.2d kN (Tension)\nFFC=%.2d kN (Tension)\nFAE=%.2d kN (Tension)",FAB,FBC,FBF,FAF,FFC,FAE)
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